Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In Griffiths's introductory quantum mechanics book, it states that if $\Psi (x,t)$ is a solution to the Schrodinger's equation, then $A\Psi (x,t)$ must also be a solution, where A is any complex constant.

This seems to suggest that for any particle with defined mass and kinetic/potential energy, there are many possible wave functions, as $A$ can be any complex constant - therefore, it gets impossible to figure out anything from the wave function.

Can anyone show me how I am mistaken?

Thank you very much.

share|improve this question
1  
I think there's something to do with normalization here--any wavefunction is fine, but a normalized wavefunction should integrate ofer all space to give 1. Not sure, though. –  Manishearth Apr 18 '12 at 16:45
    
@Manishearth: yeah, that's basically the idea. Write that up as an answer if you want. (It's possible to express this in fancier language by talking about a quotient group, but that's probably not necessary here.) –  David Z Apr 18 '12 at 17:03
2  
Except that it's the integral $\int|\psi(x,t)|^2\mathrm{d}^3x$ over all space that must be $1$, which does not determine the complex phase. However, the complex phase does not have to be determined for us to be able to determine observable probabilities. –  Peter Morgan Apr 18 '12 at 17:08
    
We can also have an infinite number of possible potential energy functions. In Newtonian Gravity, the potential energy is $V(r)=-\frac{GMm}{r}+C$, and our choice of the constant C is completely arbitrary. That doesn't mean that we can't make predictions from that potential energy function. $C$ is not observable. In much the same way, the complex phase of the wave function is not observable. –  Jerry Schirmer Apr 18 '12 at 22:58
    
@DavidZaslavsky: For some reason I didn't get pinged for that. I'm rather shaky in QM, but I tried my best to write an answer that explained it at a different level than the others. Also: I have no clue what a quotient group is :/ –  Manishearth Apr 21 '12 at 11:58
add comment

4 Answers

This is because you're not talking about a normalized wavefunction.

The wavefunction only has a physical meaning when it is interpreted as probability. The value of $\psi\bar\psi=|\psi|^2$ at a point is proportional to the probability of the wavefunction collapsing to any given point.

Since it's just proportional, we can multiply it by an arbitrary constant and not change it's physical meaning.

What we can consider is the normalised wavefunction. In this, we multiply $\psi$ by some complex constant $A$, such that the sum of all the $\psi\bar\psi$ over space is $1$--in this way, $\psi\bar\psi$ becomes the probabilty, instead of just being proportional.

In other words, we find $A$ such that $$\int (A\psi)(\bar A\bar\psi)\mathrm dV=|A|^2\int|\psi|^2\mathrm dV=1$$

Of course, you can see that we still have infinite solutions--since $A$ is a complex number, it can be any number lying on the circle of radius $\frac{1}{\sqrt{\int|\psi|^2\mathrm dV}}$ on the Argand(complex) plane.

But that's no issue--that just means that we can have a phase difference (not sure of this). Its like the phase is a normal wave--changing the phase in a normal sinusoidal wave($Asin(\omega t+\large{\phi)}$, where $\phi$ is the phase constant) doesn't change the wave, it only changes the "where the wave started", or our choice of origin. We get similar things with an integration constant.

share|improve this answer
add comment

This is the reason, why the 'correct' definition of expectation values is $$ \langle\hat T\rangle_\psi = \frac{\langle\psi | \hat T | \psi\rangle}{\langle\psi | \psi\rangle}. $$ As $\hat T$ is a linear Operator it is easy to see, that any such $A \in \mathbb{C}$ would cancel out. For normalized states, $\langle\psi | \psi\rangle = 1$, and so one recovers the usual expression $$\langle\hat T\rangle_\psi = \langle\psi | \hat T | \psi\rangle = \int \psi^*(x) \hat T \psi(x) \, dx \,.$$

share|improve this answer
add comment

I am confused by the comments. Perhaps I am missing something.

For any linear differential equation with a given solution, a constant times that solution is also a valid solution.

The Schroedinger equation is a linear differential equation for the wavefunction $\Psi$. This means that if $\Psi$ is a solution then so is $A\Psi$ where $A$ is a complex constant. You can fix this constant by further requirements like the normalization condition $\int d^dx |\Psi|^2=1$. Of course as Peter Morgan said, the phase of $A$ remains undetermined by the normalization condition.

I think this is also what Griffiths means. I quote him after he says that the wavefunction should be normalized since the particle has to be somewhere.

Well, a glance at Equation 1.1 [the Schroedinger equation] reveals that if $\Psi(x,t)$ is a solution, so too is $A\Psi(x,t)$, where $A$ is any (complex) constant.

and then he goes on to say that we must pick $A$ such that the wavefunction is normalized.

share|improve this answer
add comment

Suppose you have some operator $\hat a$ for returning some physical property, then the way you find the corresponding physical property is with the integral:

$$\int \Psi^* \hat a \Psi d^3x$$

where $\Psi^*$ is the complex conjugate. Assuming your number $A$ has modulus 1, which is required for normalisation, your integral gets multiplied by $A^*A$ and this is one. That's why multiplying by $A$ makes no difference.

The wavefunction itself cannot be physically observed. To get anything out of it requires some integral like the one above. So it's true, sort of, that the same system can be described by an infinite number of wavefunctions, but your physical predictions will always come out the same.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.